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Self Evaluation on Arithmetic Progression ICSE Class 10 Maths OP Malhotra (2026-27)

Board	ICSE
Publications	S Chand
Subject	Maths
Class	10th
Chapter-9	Arithmetic Progression
Writer	OP Malhotra
Self Evaluation	Extra Practice Questions
Edition	2026-2027

Self Evaluation on Arithmetic Progression

Self Evaluation on Arithmetic Progression ICSE Class 10 Maths OP Malhotra (2026-27)

Very Short Answer Type Questions (VSA)

1. Determine if the sequence 65, 60, 55, 50, 45, … is arithmetic. If so, give the common difference.

Sol: Given sequence = 65, 60, 55, 50, 45, …

Common difference,
d = 60 − 65 = −5
55 − 60 = −5
50 − 55 = −5
45 − 50 = −5

Since the common difference is the same throughout, the given sequence is an Arithmetic Progression (A.P.).
∴Yes, the sequence is an A.P. and its common difference is −5.

2. Find the 15th term of the sequence 5, 7, 9, 11, …

Sol: Here, a = 5, d = 2, n = 15.

Using the formula,
a_n = a + (n − 1)d

a₁₅ = 5 + (15 − 1) × 2
= 5 + 28
= 33

∴The 15th term is 33.

3. Find the next three terms of the sequence −14, −8, −2, 4, …

Sol: Common difference,
d = −8 − (−14) = 6.

Next three terms are:
4 + 6 = 10
10 + 6 = 16
16 + 6 = 22

∴The next three terms are 10, 16, 22.

4. If the first three terms of an A.P. are b, c and 2b, then find the ratio of b and c.

Sol: Since the terms are in A.P.,
c − b = 2b − c

2c = 3b

Therefore,
b : c = 2 : 3

Ans: b : c = 2 : 3.

5. Determine if the sequence 3, 6, 12, 24, 48, … is geometric. If yes, find the common ratio and the 12th term.

Sol: Common ratio,
r = 6/3 = 2
12/6 = 2
24/12 = 2

Since the common ratio is constant, the sequence is a G.P.

Here, a = 3, r = 2.

a₁₂ = ar¹¹
= 3 × 2¹¹
= 3 × 2048
= 6144

∴ Yes, it is a G.P. The common ratio is 2 and the 12th term is 6144.

6. Find the 5th term of a geometric sequence whose 3rd term is 8 and 4th term is 12.

Sol: Common ratio,
r = 12/8 = 3/2.

5th term = 4th term × r
= 12 × 3/2
= 18

∴ The 5th term is 18.

7. Find the sum of the first 100 natural numbers.

Sol: Using the formula,
S = n(n + 1)/2

S = 100 × 101 / 2
= 5050

∴ The sum of the first 100 natural numbers is 5050.

8. Find the sum of first 8 multiples of 3.

Sol: The first 8 multiples of 3 are:
3, 6, 9, 12, 15, 18, 21, 24.

Sum = 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24
= 108

∴ The required sum is 108.

9. Find the sum of n terms of the series 4 − 8 + 16 − 32 + 64 + …

Sol: Here, a = 4, r = −2.

Using the formula,
S_n = a(rⁿ − 1)/(r − 1)

S_n = 4[(-2)ⁿ − 1]/(-2 − 1)
= 4[1 − (-2)ⁿ]/3

∴ S_n = 4[1 − (−2)ⁿ]/3.

10. Find the sum to n terms of the series 1 − 3x + 9x² − 27x³ + …

Sol: Here, a = 1, r = −3x.

Using the formula,
S_n = a(rⁿ − 1)/(r − 1)

S_n = [(-3x)ⁿ − 1]/(-3x − 1)
= [1 − (−3x)ⁿ]/(1 + 3x)

∴ S_n = [1 − (−3x)ⁿ]/(1 + 3x).

Multiple Choice Questions (MCQs)

1. The 50th term of the series 5, 9, 13, 17, … is

Sol: Here, a = 5, d = 4, n = 50.

Using the formula,
a_n = a + (n − 1)d

a₅₀ = 5 + (50 − 1) × 4
= 5 + 196
= 201

Ans: (c) 201.

2. The 7th term of the series 1, 3, 9, 27, 81, … is

Sol: This is a G.P. with
a = 1, r = 3, n = 7.

Using the formula,
a_n = arⁿ⁻¹

a₇ = 1 × 3⁶
= 729

Ans: (d) 729.

3. Determine k so that k + 2, 4k − 6 and 3k − 2 are three consecutive terms of an A.P.

Sol: Since the terms are in A.P.,
2(4k − 6) = (k + 2) + (3k − 2)

8k − 12 = 4k

4k = 12

k = 3

Ans: (b) 3.

4. 19th term of the series −87, −78, −69, −60, −51, … is

Sol: Here,
a = −87, d = 9, n = 19.

Using the formula,
a₁₉ = a + (n − 1)d

a₁₉ = −87 + (19 − 1) × 9
= −87 + 162
= 75

Ans: (b) 75.

5. In an A.P. if S₂₁ = 1250, S₂₀ = 1200, find a₂₁.

Sol: The 21st term is given by,

a₂₁ = S₂₁ − S₂₀
= 1250 − 1200
= 50

Ans: (d) 50.

6. Find the sum of first 25 terms of an A.P. whose nth term is 1 − 4n.

Sol: Given,
a_n = 1 − 4n

First term,
a = 1 − 4(1) = −3

Common difference,
d = −4

Using the formula,
S_n = n/2 [2a + (n − 1)d]

S₂₅ = 25/2 [2(−3) + 24(−4)]
= 25/2 (−6 − 96)
= 25/2 (−102)
= −1275

Ans: (a) −1275.

7. Which term of the series 1/4, −1/2, 1, … is 128?

Sol: The given series is a G.P. with
a = 1/4, r = −2.

Using the formula,
a_n = arⁿ⁻¹

1/4 × (−2)ⁿ⁻¹ = 128
(−2)ⁿ⁻¹ = 512 = 2⁹

Therefore,
n − 1 = 9
n = 10

Ans: (b) 10th.

8. If 1, x, y, z, 16 are in A.P., then what is the value of x + y + z?

Sol: Here,
a = 1, l = 16, n = 5.

Common difference,
d = (16 − 1)/(5 − 1)
= 15/4

Hence,
x = 1 + 15/4 = 19/4
y = 1 + 30/4 = 17/2
z = 1 + 45/4 = 49/4

Therefore,
x + y + z = 19/4 + 17/2 + 49/4
= 102/4
= 51/2

Ans: The correct value is 51/2.
Note: None of the given options is correct. There appears to be a misprint in the question.

9. How many terms of the geometric series 1 + 4 + 16 + 64 + … will make the sum 5461?

Sol: Here,
a = 1, r = 4.

Using the formula,
S_n = (4ⁿ − 1)/3

Given,
(4ⁿ − 1)/3 = 5461
4ⁿ − 1 = 16383
4ⁿ = 16384 = 4⁷

Therefore,
n = 7

Ans: (b) 7.

10. In a G.P., the sum of the first n terms is 364, the first term is 1 and the common ratio is 3. The value of n is

Sol: Here,
a = 1, r = 3.

Using the formula,
S_n = (3ⁿ − 1)/2

Given,
(3ⁿ − 1)/2 = 364
3ⁿ − 1 = 728
3ⁿ = 729 = 3⁶

Therefore,
n = 6

Ans: (c) 6.

11. If the fifth term of a G.P. is 2, then the product of its 9 terms is

Sol: In a G.P. having an odd number of terms, the product of all the terms equals the middle term raised to the number of terms.

Middle (5th) term = 2.

Therefore,
Product = 2⁹
= 512

Ans: (b) 512.

12. In a G.P., the ratio of the sum of the first 3 terms to that of the first 6 terms is 125 : 152. Find the common ratio of the G.P.

Sol: Using the sum formula of a G.P.,

(r³ − 1)/(r⁶ − 1) = 125/152

Since,
r⁶ − 1 = (r³ − 1)(r³ + 1),
1/(r³ + 1) = 125/152
r³ + 1 = 152/125
r³ = 27/125
r = 3/5

Ans: (d) 3/5.

— : End of Self Evaluation on Arithmetic Progression ICSE Class 10 Maths OP Malhotra (2026-27) :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-10 Maths

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